differentiate x^x

AndrewTom

(a) Find $\frac{d}{dx}(x^{x})$ .

(b) Find $\frac{d}{dx}(x^{x^{x^{x^...}}})$ . For what values of $x$ is this undefined?

**Posted:** Tue Nov 17, 2009 7:44 am

nayel

a) Let $y=x^x$ . Then $\ln y=x\ln x$ , and so $\frac 1y\cdot\frac{dy}{dx}=1+\ln x$ , which implies, $\frac{dy}{dx}=x^x(1+\ln x)$ .

**Posted:** Tue Nov 17, 2009 8:53 am

_________________
Samin Riasat
Ever experienced zero gravity?

AndrewTom

Thanks nayel. I've come across the first part a number of times.For the second part I can do $x^{x^{x}}}$ and I'm thinking of building the general case up bit by bit but I'm not sure about it.

**Posted:** Tue Nov 17, 2009 11:47 am

nayel

This might be useful: let $y=x^{x^{x^{\dots}}}$ and $D_n=\frac{d}{dx}(x^{x^{x^{\dots}}})=\frac{dy}{dx}$ (there are $n$ $x$ 's). Then $D_{n+1}=\frac{d}{dx}(x^y)=\frac yx+\frac{dy}{dx}\cdot\ln x=\frac{x^{x^{x^{\dots}}}}{x}+D_n\ln x.$

**Posted:** Tue Nov 17, 2009 12:31 pm

_________________
Samin Riasat
Ever experienced zero gravity?

kcn2rivers · Hodge Conjecture Offline Joined: 05 Dec 2007 Posts: 67

Are there infinite x's in the exponent? Or do you want it for $n$ x's?
If infinite:

$y= x^{x^{x^{x...}}} = x^y$
$\ln y = y \ln x \Rightarrow \frac{dy}{dx} = \frac{y^2}{x(1-y\ln x)}$ , which is undefined for $x=0$ and $x=e^{1/e}$ .